# Question on Dirac Delta and functional derivatives

1. Jan 25, 2007

### shoehorn

Suppose that we have a compact manifold $\mathcal{M}$ with a positive definite metric $g_{ij}$. The volume of the manifold is then

$$V = \int_\mathcal{M} d^3x \sqrt{g(x)},$$

where $x^i$ are coordinates on $\mathcal{M}$ and $\sqrt{g(x)}$ is the square root of the determinant of the metric. Suppose now that we take a functional derivative of the volume with respect to the metric, i.e., we want to calculate

$$\frac{\delta}{\delta g_{ij}(y)} V = \frac{\delta}{\delta g_{ij}(y)}\int_\mathcal{M} d^3x \sqrt{g(x)}$$

As far as I know, this calculation gives

$$\frac{\delta}{\delta g_{ij}(y)} V = \frac{1}{2}\int_\mathcal{M} d^3x \sqrt{g(x)} g^{ij}(x) \delta^{(3)}(x,y)$$

Where $\delta^{(3)}(x)$ is a three-dimensional Dirac delta function. The question I have is this: is $\delta^{(3)}(x,y)$ to be regarded as (a) a unit weight density or (b) a density of weight zero?

If the answer is (a) then the result is

$$\frac{\delta}{\delta g_{ij}(y)} V = \frac{1}{2}\sqrt{g(y)} g^{ij}(y)$$

while if the answer is (b) then the result for the functional derivative is

$$\frac{\delta}{\delta g_{ij}(y)} V = \frac{1}{2} g^{ij}(y)$$

So which one is the correct answer?

Last edited: Jan 25, 2007